Question

the scatter plot shows the number of years of experience, x, and the amount charged per hour, y, for each of 25 dog sitters in new york. use the scatter plot to answer the parts below. (note that you can use the graphing tools to help you approximate the line.)
scratch area (not part of answer)
(a) write an approximate equation of the line of best fit. round the coefficients to the nearest hundredth.
( y = )
(b) using your equation from part (a), predict the amount charged per hour by a dog sitter with 10 years of experience. round your prediction to the nearest hundredth.
$

Explanation:

Step1: Find the line of best fit

To find the line of best fit, we can estimate the slope and the y - intercept. Let's pick two points on the line of best fit. Looking at the scatter plot, we can estimate two points. Let's say when \(x = 5\) (years of experience), \(y\approx8\) (amount charged per hour) and when \(x = 20\), \(y\approx20\).

The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{20 - 8}{20 - 5}=\frac{12}{15}=0.8\). Now we use the point - slope form \(y - y_1=m(x - x_1)\). Using the point \((5,8)\):

\(y - 8=0.8(x - 5)\)

\(y-8 = 0.8x-4\)

\(y=0.8x + 4\) (this is an approximate equation of the line of best fit. If we use more accurate points, for example, if we take two points from the line of best fit more precisely, let's say \((6,7)\) and \((18,19)\), the slope \(m=\frac{19 - 7}{18 - 6}=\frac{12}{12} = 1\), and using point - slope with \((6,7)\): \(y - 7=1\times(x - 6)\), so \(y=x + 1\). But let's assume we use the first approximation for now. Wait, maybe a better way is to use the trend. Let's re - evaluate. Let's take two points on the line of best fit. Looking at the scatter plot, when \(x = 0\), the y - intercept is around \(y = 6\) (approximate), and when \(x = 20\), \(y\approx22\). Then the slope \(m=\frac{22 - 6}{20-0}=\frac{16}{20}=0.8\). So the equation is \(y = 0.8x+6\)? Wait, maybe my initial point selection was wrong. Let's do it more carefully. Let's take two points that are on the line of best fit. Let's say one point is \((5,8)\) and another is \((15,16)\). Then the slope \(m=\frac{16 - 8}{15 - 5}=\frac{8}{10}=0.8\). Then using \(y=mx + b\), plugging in \((5,8)\): \(8=0.8\times5 + b\), \(8 = 4 + b\), so \(b = 4\). So the equation is \(y=0.8x + 4\).

Step2: Predict for \(x = 10\)

Now, we use the equation from part (a) \(y = 0.8x+4\) to predict the amount charged per hour when \(x = 10\) (10 years of experience).

Substitute \(x = 10\) into the equation:

\(y=0.8\times10 + 4\)

\(y = 8+4\)

\(y = 12\) (If we used a different line of best fit, say \(y=x + 1\), then \(y=10 + 1=11\), but our first approximation gives \(y = 12\). However, if we look at the scatter plot, when \(x = 10\), the points are around \(y = 10 - 12\). Let's assume our line of best fit is \(y = 0.8x+4\), then for \(x = 10\), \(y=0.8\times10 + 4=12\).

Answer:

(a) An approximate equation of the line of best fit is \(y = 0.8x+4\) (answers may vary slightly depending on the points chosen for the line of best fit). (b) The predicted amount charged per hour for a dog sitter with 10 years of experience is \(\$12\) (or a value close to it depending on the line of best fit equation).